The Stacks project

Proof. Assume $\mathcal{Z} \subset \mathcal{Y}$ is a closed substack and $f \circ g$ factors through $\mathcal{Z}$. To prove the lemma it suffices to show that $f$ factors through $\mathcal{Z}$. Consider a scheme $T$ and a morphism $T \to \mathcal{X}$ given by an object $x$ of the fibre category of $\mathcal{X}$ over $T$. We will show that $f(x)$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the projection $T \times _\mathcal {X} \mathcal{W} \to T$ is a surjective, flat, locally finitely presented morphism. Hence there is an fppf covering $\{ T_ i \to T\}$ such that $T_ i \to T$ factors through $T \times _\mathcal {X} \mathcal{W} \to T$ for all $i$. Then $T_ i \to \mathcal{X}$ factors through $\mathcal{W}$ and hence $T_ i \to \mathcal{Y}$ factors through $\mathcal{Z}$. Thus $x|_{T_ i}$ is an object of $\mathcal{Z}$. Since $\mathcal{Z}$ is a strictly full substack, we conclude that $x$ is an object of $\mathcal{Z}$ as desired. $\square$

Proof. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. By Lemma 101.38.2 we may replace $\mathcal{Y}$ by $V$. Thus it suffices to show that if $X \to \mathcal{X}$ is a morphism from a scheme to an algebraic stack, then the scheme theoretic image exists. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Set $R = U \times _\mathcal {X} U$. We have $\mathcal{X} = [U/R]$ by Algebraic Stacks, Lemma 94.16.2. By Properties of Stacks, Lemma 100.9.11 the closed substacks $\mathcal{Z}$ of $\mathcal{X}$ are in $1$-to-$1$ correspondence with $R$-invariant closed subschemes $Z \subset U$. Let $Z_1 \subset U$ be the scheme theoretic image of $X \times _\mathcal {X} U \to U$. Observe that $X \to \mathcal{X}$ factors through $\mathcal{Z}$ if and only if $X \times _\mathcal {X} U \to U$ factors through the corresponding $R$-invariant closed subscheme $Z$ (details omitted; hint: this follows because $X \times _\mathcal {X} U \to X$ is surjective and smooth). Thus we have to show that there exists a smallest $R$-invariant closed subscheme $Z \subset U$ containing $Z_1$.

Let $\mathcal{I}_1 \subset \mathcal{O}_ U$ be the quasi-coherent ideal sheaf corresponding to the closed subscheme $Z_1 \subset U$. Let $Z_\alpha$, $\alpha \in A$ be the set of all $R$-invariant closed subschemes of $U$ containing $Z_1$. For $\alpha \in A$, let $\mathcal{I}_\alpha \subset \mathcal{O}_ U$ be the quasi-coherent ideal sheaf corresponding to the closed subscheme $Z_\alpha \subset U$. The containment $Z_1 \subset Z_\alpha$ means $\mathcal{I}_\alpha \subset \mathcal{I}_1$. The $R$-invariance of $Z_\alpha$ means that

as (quasi-coherent) ideal sheaves on (the algebraic space) $R$. Consider the image

Since direct sums of quasi-coherent sheaves are quasi-coherent and since images of maps between quasi-coherent sheaves are quasi-coherent, we find that $\mathcal{I}$ is quasi-coherent. Since pull back is exact and commutes with direct sums we find

Hence $\mathcal{I}$ defines an $R$-invariant closed subscheme $Z \subset U$ which is contained in every $Z_\alpha$ and contains $Z_1$ as desired. $\square$

Proof. The scheme theoretic inverse image of $\mathcal{Z}_2$ in $\mathcal{Y}_1$ is a closed substack of $\mathcal{Y}_1$ through which $f_1$ factors. Hence $\mathcal{Z}_1$ is contained in this. This proves the lemma. $\square$

Proof. Let $\mathcal{Y}' \to \mathcal{Y}$ be a flat morphism of algebraic stacks. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose a scheme $V'$ and a surjective smooth morphism $V' \to \mathcal{Y}' \times _\mathcal {Y} V$. We may and do assume that $V = \coprod _{i \in I} V_ i$ is a disjoint union of affine schemes and that $V' = \coprod _{i \in I} \coprod _{j \in J_ i} V_{i, j}$ is a disjoint union of affine schemes with each $V_{i, j}$ mapping into $V_ i$. Let

1. $\mathcal{Z} \subset \mathcal{Y}$ be the scheme theoretic image of $f$,

2. $\mathcal{Z}' \subset \mathcal{Y}'$ be the scheme theoretic image of the base change of $f$ by $\mathcal{Y}' \to \mathcal{Y}$,

3. $Z \subset V$ be the scheme theoretic image of the base change of $f$ by $V \to \mathcal{Y}$,

4. $Z' \subset V'$ be the scheme theoretic image of the base change of $f$ by $V' \to \mathcal{Y}$.

If we can show that (a) $Z = V \times _\mathcal {Y} \mathcal{Z}$, (b) $Z' = V' \times _{\mathcal{Y}'} \mathcal{Z}'$, and (c) $Z' = V' \times _ V Z$ then the lemma follows: the inclusion $\mathcal{Z}' \to \mathcal{Z} \times _\mathcal {Y} \mathcal{Y}'$ (Lemma 101.38.4) has to be an isomorphism because after base change by the surjective smooth morphism $V' \to \mathcal{Y}'$ it is.

Proof of (a). Set $R = V \times _\mathcal {Y} V$. By Properties of Stacks, Lemma 100.9.11 the rule $\mathcal{Z} \mapsto \mathcal{Z} \times _\mathcal {Y} V$ defines a $1$-to-$1$ correspondence between closed substacks of $\mathcal{Y}$ and $R$-invariant closed subspaces of $V$. Moreover, $f : \mathcal{X} \to \mathcal{Y}$ factors through $\mathcal{Z}$ if and only if the base change $g : \mathcal{X} \times _\mathcal {Y} V \to V$ factors through $\mathcal{Z} \times _\mathcal {Y} V$. We claim: the scheme theoretic image $Z \subset V$ of $g$ is $R$-invariant. The claim implies (a) by what we just said.

For each $i$ the morphism $\mathcal{X} \times _\mathcal {Y} V_ i \to V_ i$ is quasi-compact and hence $\mathcal{X} \times _\mathcal {Y} V_ i$ is quasi-compact. Thus we can choose an affine scheme $W_ i$ and a surjective smooth morphism $W_ i \to \mathcal{X} \times _\mathcal {Y} V_ i$. Observe that $W = \coprod W_ i$ is a scheme endowed with a smooth and surjective morphism $W \to \mathcal{X} \times _\mathcal {Y} V$ such that the composition $W \to V$ with $g$ is quasi-compact. Let $Z \to V$ be the scheme theoretic image of $W \to V$, see Morphisms, Section 29.6 and Morphisms of Spaces, Section 67.16. It follows from Lemma 101.38.2 that $Z \subset V$ is the scheme theoretic image of $g$. To show that $Z$ is $R$-invariant we claim that both

are the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} R \to R$. Namely, we first use Morphisms of Spaces, Lemma 67.30.12 to see that $\text{pr}_0^{-1}(Z)$ is the scheme theoretic image of the composition

Since the first arrow here is surjective and smooth we see that $\text{pr}_0^{-1}(Z)$ is the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} R \to R$. The same argument applies that $\text{pr}_1^{-1}(Z)$. Hence $Z$ is $R$-invariant.

Statement (b) is proved in exactly the same way as one proves (a).

Proof of (c). Let $Z_ i \subset V_ i$ be the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V_ i \to V_ i$ and let $Z_{i, j} \subset V_{i, j}$ be the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V_{i, j} \to V_{i, j}$. Clearly it suffices to show that the inverse image of $Z_ i$ in $V_{i, j}$ is $Z_{i, j}$. Above we've seen that $Z_ i$ is the scheme theoretic image of $W_ i \to V_ i$ and by the same token $Z_{i, j}$ is the scheme theoretic image of $W_ i \times _{V_ i} V_{i, j} \to V_{i, j}$. Hence the equality follows from the case of schemes (Morphisms, Lemma 29.25.16) and the fact that $V_{i, j} \to V_ i$ is flat. $\square$

Proof. Let $z \in |\mathcal{Z}|$ be a point. Choose an affine scheme $V$, a point $v \in V$, and a smooth morphism $V \to \mathcal{Y}$ mapping $v$ to $z$. Then $\mathcal{X} \times _\mathcal {Y} V$ is a quasi-compact algebraic stack. Hence we can find an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X} \times _\mathcal {Y} V$. By Lemma 101.38.5 the scheme theoretic image of $\mathcal{X} \times _\mathcal {Y} V \to V$ is $Z = \mathcal{Z} \times _\mathcal {Y} V$. Hence the inverse image of $|\mathcal{Z}|$ in $|V|$ is $|Z|$ by Properties of Stacks, Lemma 100.4.3. By Lemma 101.38.2 $Z$ is the scheme theoretic image of $W \to V$. By Morphisms of Spaces, Lemma 67.16.3 we see that the image of $|W| \to |Z|$ is dense. Hence the image of $|\mathcal{X} \times _\mathcal {Y} V| \to |Z|$ is dense. Observe that $v \in Z$. Since $|V| \to |\mathcal{Y}|$ is open, a topology argument tells us that $z$ is in the closure of the image of $|f|$ as desired. $\square$

Proof. By Lemma 101.7.7 the morphism $s : \mathcal{V} \to \mathcal{Y}$ is quasi-compact. Hence the construction of the scheme theoretic image $\mathcal{Y}'$ of $s$ commutes with flat base change by Lemma 101.38.5. Thus to prove the lemma we may assume $\mathcal{Y}$ is representable by an algebraic space and we reduce to the case of algebraic spaces which is Morphisms of Spaces, Lemma 67.16.7. $\square$